Message/Author |
|
cricket posted on Monday, May 24, 2004 - 8:59 am
|
|
|
When I run mixture models with all available data (e.g., type=mixture missing;), I get a likelihood ratio chi-square test for MCAR. However, I don't get that test when running ordinary LGM with (type = meanstructure missing h1). Is there any way to get that same test? |
|
bmuthen posted on Monday, May 24, 2004 - 9:57 am
|
|
|
You only get this test with categorical outcomes. |
|
cricket posted on Monday, May 24, 2004 - 10:07 am
|
|
|
I ran an LGM with my categorical outcome with the same instructions as above (i.e., analysis: meanstructure missing h1;), and I still didn't get the test. |
|
|
I should have said you only get this test with categorical outcomes and TYPE=MIXTURE not with an ordinary non-mixture growth model. You could use TYPE=MIXTURE and CLASSES = c (1); if you really want it. |
|
cricket posted on Monday, May 24, 2004 - 10:26 am
|
|
|
Thanks |
|
|
Is there information on the way that MPlus calculates its chi-square test of MCAR. I have looked through the manual and technical appendices, perhaps I have missed it. |
|
|
See Technical Appendix 6 formula 133. |
|
Jon Elhai posted on Sunday, June 15, 2008 - 9:12 am
|
|
|
Drs. Muthen, What is the command syntax for obtaining Little's MCAR test when using type=mixture? |
|
|
This is available only for categorical outcomes and is given automatically in this case. |
|
Tracy Witte posted on Wednesday, November 12, 2008 - 11:37 am
|
|
|
For the chi-square test for MCAR, I get a value of 111.37, df = 222, p = 1.0000 for the pearson & 85.65, df = 222, & p = 1.0000 for the likelihood ratio. Does this mean that MCAR does hold? Or do the p values of 1 mean that this test is uninterpretable? |
|
|
The discrepancy between the 2 chi-2's probably indicate that neither is trustworthy due to many zero cells. |
|
Tracy Witte posted on Wednesday, November 12, 2008 - 1:42 pm
|
|
|
So should I just assume MAR to be safe? |
|
|
Yes. |
|
Aleksandra posted on Monday, December 08, 2008 - 8:34 pm
|
|
|
Dr. Muthien, what is the cost of specifying the data as MAR when the data are actually MCAR? Thank you very much. Alex |
|
|
There is no cost. When MCAR holds, MAR holds. |
|
Wen-Hsu Lin posted on Sunday, October 19, 2014 - 7:38 pm
|
|
|
Thank you. One additional question of setting up the imputation is that only integer is allowed. So, should I set rounding = 0;? Thanks a lot. |
|
|
If you want no decimals, specify ROUNDING=0; |
|
|
Hello, I am running a growth model using four time points (observed scores across four waves) and am trying to assess whether the data meets the Little's MCAR assumption. The composite scores are categorical (with scores of 0, 1, or 2). Using MLR, I obtain the following results for the Little's MCAR. Pearson Chi Squared: 190.027 Pearson df: 160 Pearson p: 0.0526 Likelihood Chi Squared: 207.696 Likelihood df: 160 Likelihood p: 0.0066 Because my data are categorical and I am seeking fit statistics, I will be using WLSMV estimation for my main analyses. Thus, can I trust the Pearson Chi Square Little's MCAR results and assume my data is missing completely at random? Thanks! Hillary |
|
|
I would recommend to make sure that 1. estimator=wlsmv; parameterization=theta; gives similar results to 2. estimator=ml; link=probit; and 3. estimator=bayes; as evidence that the MCAR assumption is not violated. The PPP value in Bayes can also be used as a fit index. The MCAR test with large number of cells needs large amount of data for proper asymptotic conclusion. Because of the different conclusions you have obtained we can see that the data is not sufficient to get the proper asymptotic behavior anyway. Obviously however there is no strong evidence for MCAR violation and the WLSMV results are most likely fine. |
|
|
Thank you for these suggestions!! Results look very similar. Hillary |
|
Back to top |